A simple battery eliminator circuit is a clever device that provides a steady direct current voltage to different electronic gadgets, so you do not need to use a regular battery.
This kind of circuit is especially useful for low power devices that need a dependable and stable voltage source.
In this article we will explore how to design and use a simple battery eliminator circuit that uses the LM317 adjustable voltage regulator integrated circuit.
We will also in our circuit diagram of a battery eliminator circuit that can replace 9V batteries.
This setup can be used to power any device that works with a 9V battery.
Circuit Working:
Parts List:
| Component | Specification | Quantity |
|---|---|---|
| Resistors | ||
| 1.5k | 1/4 watt | 1 |
| 220Ω | 1/4 watt | 1 |
| 15k | 1/4 watt | 1 |
| Capacitors | ||
| Electrolytic | 10µF 25V | 2 |
| Semiconductors | ||
| IC | LM317 | 1 |
| Diodes | 1N4007 for bridge rectifier | 4 |
| Transformer | 230V primary, 9V secondary 1.5A | 1 |
In this article transformer takes the 230V alternating current from the power supply and changes it to a lower voltage of 9V alternating current.
After that, this alternating current is converted into pulsating direct current using a bridge rectifier made of four diodes each capable of handling 1 ampere.
To improve the quality of this pulsating direct current a capacitor C1 which has a capacitance of 10μF is used as a filter.
The cleaned up direct current then goes into the IC LM317 voltage regulator chip, which provides a steady and adjustable output voltage.
The output voltage depends on the values of resistors R1 and R3 which are important for setting this voltage.
The another capacitor C2 with a same capacitance of 10μF placed at the output to further smooth out the regulated direct current voltage making sure any ripples are kept to a minimum.
Formulas with Calculations:
Formulas with calculations for Simple Battery Eliminator Circuit using IC LM317 is mentioned below:
Output Voltage Formula (LM317):
The output voltage (Vout) of the LM317 is given by:
Vout = Vref × (1 + R2 / R1) + Iadj × R2
where,
- Vref is 1.25V reference voltage of IC LM317
- Iadj adjustment pin current is typically very small ~50µA and often neglected.
Simplified formula:
Vout = Vref × (1 + R2 / R1)
Component Values:
R1 = 1.5k = 1500Ω
R2 = 220Ω
R3 = 15k = 15,000Ω (optional resistor for fine tuning)
Vref = 1.25V
Calculations for Vout:
Using the formula:
Vout = 1.25 × (1 + R2 / R1)
Substitute the values:
Vout = 1.25 × (1 + 220 / 1500)
Vout = 1.25 × (1 + 0.1467)
Vout = 1.25 × 1.1467
Vout = 1.43 V
If R3 is included for fine tuning the resistance values will need to be combined and the output voltage may vary accordingly.
How to Build:
To build a Simple Battery Eliminator Circuit using IC LM317 follow the below mentioned steps:
- Assemble all the components mentioned in the above circuit diagram
- Connect positive of capacitor C1 from input pin of IC1 and negative of C1 to GND.
- Connect ADJ pin of IC1 to one end of resistor R1 and other end of resistor R1 to GND.
- Connect the ADJ pin of IC1 between resistors R2 and R3
- Connect the output pin of IC1 to positive of Vout 9V DC supply and negative to the GND.
- Connect resistor R2 and resistor R3 from output pin of IC1 and GND.
- Connect positive of capacitor C2 from output pin of IC1 and negative of C1 to GND.
- Connect one pin of bridge rectifier to input pin of IC1 other pin of bridge rectifier to GND third pin of bridge rectifier to one wire of 9V transformer and fourth pin pf bridge rectifier to 2nd wire of 9V transformer.
Conclusion:
The Simple Battery Eliminator Circuit using IC LM317 we are talking about is an easy and efficient way to get a steady DC voltage.
It is perfect for different electronic projects where using a battery is not a good option.
The LM317 integrated circuit helps keep the voltage consistent and because the circuit is so simple, it is great for beginners and hobbyists.